Question 20. Let A be a square matrix such that AA^{T}=I.
Then, \frac{1}{2}A\left[(A+A^{T})^{2}+(A-A^{T})^{2}\right] is equal to
(1) A^{2}+I
(2) A^{3}+I
(3) A^{2}+A^{T}
(4) A^{3}+A^{T}
Answer (4)
Explanation:
Given, AA^{T}=I.
This also implies, A^{T}A=I.
Expand the given expression.
\frac{1}{2}A\left[A^{2}+(A^{T})^{2}+2AA^{T}+A^{2}+(A^{T})^{2}-2AA^{T}\right].
The terms 2AA^{T} cancel out.
So we get, \frac{1}{2}A\left[2A^{2}+2(A^{T})^{2}\right].
Simplifying, A\left[A^{2}+(A^{T})^{2}\right].
Now, A(A^{T})^{2}=A^{T}.
Hence, A^{3}+A^{T}.
Therefore, option (4) is correct.